The base of a solid is the region enclosed by the graphs of $y=x^2-5x+7$ and $y=3$, between $x=1$ and $x=4$. Cross sections of the solid perpendicular to the $x$ -axis are rectangles whose height is $x$. Which one of the definite integrals gives the volume of the solid? Choose 1 answer: Choose 1 answer: (Choice A) A $\int_1^4 (x^2-5x+4)\cdot x\,dx$ (Choice B) B $\int_1^4 (x^2-5x+4)^2\,dx$ (Choice C) C $\int_1^4 (-x^2+5x-4)\cdot x\,dx$ (Choice D) D $\pi\int_1^4 (x^2-5x+4)^2\,dx$
Solution: It's always a good idea to graph the situation before we solve. $y$ $x$ $ R$ ${y=x^2-5x+7}$ ${y=3}$ $(1,3)$ $(4,3)$ Let's imagine the solid is made out of many thin slices. $y$ $x$ Each slice is a prism. Let the width of each slice be $dx$ and let the area of the prism's face, as a function of $x$, be $A(x)$. Then, the volume of each slice is $A(x)\,dx$, and we can sum the volumes of infinitely many such slices with an infinitely small width using a definite integral: $\int_a^b A(x)\,dx$ What we now need is to figure out the expression of $A(x)$ and the interval of integration. Let's consider one such slice. $y$ $x$ ${y=x^2-5x+7}$ ${y=3}$ $(1,3)$ $(4,3)$ $ b(x)$ $ h(x)$ $ dx$ $ A(x)$ The face of that slice is a rectangle with base $b(x)$ and height $h(x)$. We are given that $h(x)=x$, and we know that for each value of $x$, the base $b(x)$ is equal to the difference between ${y=3}$ and ${y=x^2-5x+7}$. Now we can find an expression for the area of the face of the prism: $\begin{aligned} &\phantom{=}A(x) \\\\ &=b(x)h(x) \\\\ &=[{3}-({x^2-5x+7})]\cdot x \\\\ &=(-x^2+5x-4)\cdot x \end{aligned}$ The leftmost endpoint of the base of the solid is at $x=1$ and the rightmost endpoint is at $x=4$. So the interval of integration is $[1,4]$. Now we can express the definite integral in its entirety! $\int_1^4 (-x^2+5x-4)\cdot x\,dx$